来几个RF问题，看看大家的RF基础功底——系列一

checkz

RF Quiz #1
1. What is the impedance of free space?

2. What happens to the noise figure of a receiver when a 10 dB attenuator is added at the input?
a) Noise figure increases by 10 dB
b) Noise figure decreases by 10 dB
c) Noise figure doesn't change
3. An RF system has a linear throughput gain of +10 dB and an output 3rd-order intercept point (OIP3) of +30 dBm. What is the input 3rd-order intercept point (IIP3)?
a) +20 dBm
b) +40 dBm
c) +30 dBm
4. Which filter type has the greatest selectivity for a given order (i.e., N=5)?
a) Bessel
b) Chebychev (ripple=0.1 dB)
c) Butterworth
5. Which mixer spurious product is a 5th-order product?
a) 1*LO + 5*RF
b) 6*LO - 1*RF
c) 3*LO - 2*IF
6. A 2.8 GHz oscillator is phase-locked to a 10 MHz reference oscillator that has a single-sided phase noise of -100 dBc at 1 kHz offset. What is the single-sided phase noise of the 2.8 GHz oscillator at 1 kHz offset?
a) -48.6 dBc
b) -100 dBc
c) -51.1 dBc
7. What is the power of a 2 Vpk-pk sine wave across a 50 ohm load?
a) -20.0 dBm
b) +10.0 dBm
c) +19.0 dBm
8. Which 2-port S-parameter is commonly referred to as "reverse isolation" in an amplifier?
a) S21
b) S22
c) S12
9. What are the minimum and maximum combined VSWR limits at an interface characterized by a 1.25:1 VSWR and a 2.00:1 VSWR?
a) 1.75:1 (min), 2.25:1 (max)
b) 1.60:1 (min), 2.50:1 (max)
c) 0.75:1 (min), 3.25:1 (max)
10. An ideal directional coupler has a directivity of 25 dB and an isolation of 40 dB. What is its coupling value?
a) 65 dB
b) 40 dB
c) 15 dB

好贴！置顶几天，大家都看看自己还有哪不足

[此贴子已经被awp666于2006-6-9 15:43:55编辑过]

bestdesign

1,277 or 300ohm
2,c
3,a
4,??
5,c
6,c
7,b
8,c
9,??
10,c
对于我来说，好难，谢谢checkz.

checkz

要下班了，明天上传答案和系列二

bestdesign

好啊，期待着指教。

checkz

Answer to Quiz #1
1. What is the impedance of free space?
b) 120 p (376.991) ohms
Impedance is defined as the square root of the ratio of the permeability (mu, m) to the permittivity (epsilon, e) - in this case of free space.
e0 = 1/(36 p) * 10-9 F/m
m0 = 4 p * 10-7  H/m
z0 = sqrt [m0 / e0] = sqrt [4 p * 10-7 * 36 p * 109] = sqrt [144 * 102 * p2] = 120 p = 376.991 (377) W.
2. What happens to the noise figure of a receiver when a 10 dB attenuator is added at the input?
a) Noise figure increases by 10 dB
The formula for cascaded noise figure is:
NF(M stages) = 10*log [nf1 + (nf2-1)/(gain1) + (nf3-1)/(gain1*gain2) + ... + (nfM-1)/(gain1*gain2*...*gainM-1)];
where each "nf" and "gain" value is expressed as a ratio rather than in dB, and M is the total number of stages.
The noise figure of an attenuator is equal to its insertion loss (10 dB in this case). Note that per the equation that the noise figure of the first element in the chain is not modified by the gain of preceding stages - as are the subsequent stages' noise figures. Therefore, any noise figure added to the front end adds directly to the overall system noise figure - in this case an increase of 10 dB.
3. An RF system has a linear throughput gain of +10 dB and an output 3rd-order intercept point (OIP3) of +30 dBm. What is the input 3rd-order intercept point (IIP3)?
a) +20 dBm
A system's 3rd-order intercept points are determined by the components between its input and output. As with the input signal power, the system's gain modifies the output intercept point values. Simply add the gain to the IIP3 to arrive at the OIP3.
+30 dBm - 10 dB = +20 dBm.
4. Which filter type has the greatest selectivity for a given order (i.e., N=5)?
b) Chebychev (ripple=0.1 dB)
Many texts exist that list the transfer functions of the major filter types. Rather than attempt to reiterate them all here, the following list presents them in order of increasing selectivity. The price to be paid for increased selectivity is a greater slope in the group delay near the band edges (bad for digital communications).
1. Bessel (Bessel-Thompson)
2. Gaussian (only slightly greater than Bessel)
3. Butterworth
4. Chebychev
5. Elliptic (a.k.a. Cauer-Chebychev, or CC)
5. Which mixer spurious product is a 5th-order product?
c) 3*LO - 2*IF
The order of any product is (±j*LO ±k*IF) simply the sum of the harmonic orders of the two signals that create it. In this example, the 3rd harmonic of the LO (local oscillator) and the 2nd harmonic of the IF (intermediate frequency) combine to generate the 5th-order product. The mathematical sign of the operation does not affect the order so that Order = | j | + | k |.
Order = | 3 | + | -2 | = 5.
6. A 2.8 GHz oscillator is phase-locked to a 10 MHz reference oscillator that has a single-sided phase noise of -100 dBc at 1 kHz offset. What is the single-sided phase noise of the 2.8 GHz oscillator at 1 kHz offset?
a) -48.6 dBc
When an oscillator (2.8 GHz in this case) is phase-locked (PLO) to a reference source (10 MHz in this case), the phase noise is increased in amplitude by an amount equal to 20*log(fPLO/fRef) + 2.5 dB, where the additional 2.5 dB (rule of thumb) is due to phase noise added by the phase locking circuitry. This explains why an extremely low phase noise reference oscillator is required when being used with a microwave frequency PLO.
-100 dBc + [20*log (2800/10) + 2.5] dB = -48.6 dBc.
7. What is the power of a 2 Vpk-pk sine wave across a 50 ohm load?
b) +10.0 dBm
Power in a sine wave is based on its rms voltage and the impedance it is imposed across. The rms value of a sine wave's peak-to-peak value is equal to its peak value divided by the square root of two. Rather than trying to remember whether to use 20*log or 10*log for conversion to decibels (a real stumbling point for a lot of people), just remember to always use 10* log when dealing with power and calculate power first using Ohm's Law (V^2 / R). Since the answer will be in watts, you'll need to either convert to milliwatts prior to converting to dBm, or add 30 dB to the result.
Vrms = 2 / 2 / sqrt (2) = 0.7071
P = sqr (0.7071) / 50 = 0.01 W (0.01 * 1000 = 10 mW)
10*log (0.01) = -20 dBW + 30 dB = +10 dBm
10*log (0.01 * 1000) = +10 dBm.
8. Which 2-port S-parameter is commonly referred to as "reverse isolation" in an amplifier?
c) S12
Common names for each of the four 2-port S-parameters are:
S11 : input return loss
S21 : forward gain
S12 : reverse isolation (or reverse gain)
S22 : output return loss
9. What are the minimum and maximum combined VSWR limits at an interface characterized by a 1.25:1 VSWR and a 2.00:1 VSWR?
b) 1.60:1 (min), 2.50:1 (max)
When a signal interface is composed of two elements with differing complex impedances, part (maybe all) of the incident signal will be reflected. Since VSWR is a scalar value, the phase information of the reflection coefficient is lost in the conversion. Therefore a best case and a worst case total combined VSWR  is calculated as follows:
VSWR (max) = [ VSWR1 * VSWR2 ] : 1
VSWR (min) = [ VSWR1 / VSWR2 ] : 1, where VSWR1 > VSWR2.
VSWR (max) = [2.00 * 1.25] : 1 = 2.50 : 1
VSWR (min) = [2.00 / 1.25] : 1 = 1.60 : 1.
10. An ideal directional coupler has a directivity of 25 dB and an isolation of 40 dB. What is its coupling value?
c) 15 dB
A directional coupler is characterized by five main parameters as follows:
1.  Frequency band of operation.
2.  Power coupling expressed as dB down from the input power level.
3.  Isolation of the coupled port from the output port (essentially coupling factor from the output port to the coupled port)
4.  Directivity, which  is mathematically the difference between the magnitudes of the isolation and the coupling. If the coupler in this case had 0 dBm signals applied to both the input and output ports, the coupled port would see -15 dBm from the input port and -40 dB from the output port, hence, an isolation of 25 dB.
Coupling = 40 dB - 25 dB = 15 dB.

checkz

RF Quiz #2
1. On a Smith chart, what does a point in the bottom half of the chart represent?
a) An inductive impedance
b) A capacitive impedance
c) Power saturation
2. While we're on the subject of Smith charts, what is the impedance of the point at the far left edge of the center horizontal line?
a) Infinite ohms (open circuit)
b) Zero ohms (short circuit)
c) 50 ohm match
3. A single-conversion downconverter uses a high-side local oscillator (LO) to translate the input radio frequency (RF) to an intermediate frequency (IF). Will spectral inversion occur at IF?
a) Yes, always
b) No, never
c) Sometimes
4. What happens to the noise floor of a spectrum analyzer when the input filter resolution bandwidth is decreased by two decades?
a) 20 dB increase
b) 20 dB decrease
c) 40 dB decrease
5. What is a primary advantage of a quadrature modulator?
a) Low LO power required
b) Four separate outputs
c) Single-sideband output
6. What is meant by dBi as applied to antennas?
a) Isolation in decibels
b) Physical size relative to intrinsic antennas
c) Gain relative to an isotropic radiator
7. What is the power dynamic range of an ideal 12-bit analog-to-digital converter (ADC)?
a) 36.12 dB
b) 120 dB
c) 72.25 dB
8. An ideal 10 dB attenuator is added in front of a load that has a 2.00:1 VSWR. What is the resulting VSWR of the load + attenuator?
a) 1.07:1
b) 2.10:1
c) 12.0:1
9. What is the thermal noise power in a 1 MHz bandwidth when the system temperature is 15 degrees Celsius (assume gain and noise figure are 0 dB)?
a) -114.0 dBm (in a 1 MHz bandwidth)
b) -114.0 dBm
c) -114.0 dBm/Hz
10. Two equal amplitude tones have a power of +10 dBm, and generate a pair of equal amplitude 3rd-order intermodulation products at -20 dBm. What is the 2-tone, 3rd-order intercept point (IP3) of the system?
a) +40 dBm
b) +25 dBm
c) +20 dBm

Julie

<P>系列二1 b  2 b 6c</P>

bestdesign

惭愧，太惭愧了。

checkz

答案很简单，其中有好多更深层次的理论知识需要挖掘，希望大家踊跃参加讨论，相信可以学到更多。

checkz

<P>系列一的第九题没看懂啥意思，有高手能讲解一下吗？</P><P>是两个分别具有1.25:1和2:1的模块串连吗？</P>

checkz

Answer to Quiz #2
1. On a Smith chart, what does a point in the bottom half of the chart represent?
b) A capacitive impedance
Points in the bottom half of the Smith chart represent capacitive impedances while points in the top half represent inductive impedances. Both cases include a resistive component, also. Points that lie along the center horizontal represent pure resistances.
2. While we're on the subject of Smith charts, what is the impedance of the point at the far left edge of the center horizontal line?
b) Zero ohms (short circuit)
The Smith chart's bordering circle is the locus of points whose reflection coefficients are of magnitude one. Here are a few of the major points on the Smith chart (50 W system):
1.  Left center : short circuit (0 ± j0 W).
2.  Top center : pure inductive reactance (0 + j50 W).
3.  Right center : open circuit (0 ± j&yen; W).
4.  Bottom center : pure capacitive reactance (0 - j50 W).
5.  Dead center : pure 50 ohms (50 ± j0 W).
3. A single-conversion downconverter uses a high-side local oscillator (LO) to translate the input radio frequency (RF) to an intermediate frequency (IF). Will spectral inversion occur at IF?
a) Yes, always
Spectral inversion occurs when high frequencies within the input signal bandwidth are translated to low frequencies in the output bandwidth, and vice versa. Since a downconversion is being performed, the lower sideband of the mixing process is extracted, hence the difference between the LO frequency and the RF frequency is desired. Consider the following parameters and how spectral inversion occurs.
RF input frequency band : fc = 1250 MHz, BW = 100 MHz (1200 - 1300 MHz).
LO frequency : 1600 MHz.
IF output frequency band : fc = 350 MHz, BW = 100 MHz (300 - 400 MHz).
When the lower frequency of the input band is subtracted from the LO frequency (1600 MHz - 1200 MHz = 400 MHz) a larger frequency is obtained than when the higher frequency of the input band is subtracted from the LO frequency (1600 MHz - 1300 MHz = 300 MHz). This means that the output spectrum is the mirror image of the input spectrum.
How to avoid spectral inversion? Always use a low-side LO (LO frequency below RF input frequency band) for mixing, or ensure that an even number of spectral inversions are performed in the converter (i.e., two stages of conversion with high-side LO's).
4. What happens to the noise floor of a spectrum analyzer when the input filter resolution bandwidth is decreased by two decades?
b) 20 dB decrease
The input filter bandwidth determines the amount of power that will be present at the detector circuitry. Since the detector performs a power integration function, it sums all of the incident power across the band. Decreasing the bandwidth by a factor of 100 (two decades) allows one one-hundredth of the amount of power to reach the detector, which in term of decibels is:
10*log( 1/100) = -20 dB.
5. What is a primary advantage of a quadrature modulator?
c) Single-sideband output
A quadrature modulator is comprised of two mixers, each of which receives input data and local oscillator (LO) signals that are shifted 90 degrees relative to each other. The outputs are summed together to generate the single-sideband signal. Deviation of the phases from the ideal 90 degrees and deviations from equal amplitudes going into the mixers will result in less than perfect undesired sideband cancellation. Which sideband gets canceled depends on the phase relationship of the signals entering the mixers.
                The two mixer outputs are:
m1 (t) = cos (wL*t) * cos (wI*t) = 1/2 * cos (wL*t - wI*t) + 1/2 *cos (wL*t + wI*t)
m2 (t) = cos (wL*t - pi/2) * cos (wI*t -pi/2) = sin (wL*t) * sin (wI*t)
         = 1/2 * cos (wL*t - wI*t) - 1/2 *cos (wL*t + wI*t)
Now sum the m1 (t) and m2 (t) outputs:
f (t) = 1/2 * cos (wL*t - wI*t) + 1/2 *cos (wL*t + wI*t) + 1/2 * cos (wL*t - wI*t)
                                                                             - 1/2 *cos (wL*t + wI*t)
f (t) = cos (wL*t - wI*t)
Note that what remains is the lower sideband. Upper sideband cancellation can be achieved by rearranging the 90 degree power splitters. If the data input is digital, the data streams can be digitally shifted by 90 degrees and the first 90 degree power splitter can be eliminated.
6. What is meant by dBi as applied to antennas?
c) Gain relative to an isotropic radiator
An isotropic radiator (antenna) emits electromagnetic energy equally in all directions as if it were originating from a point source. Equipotential surfaces are spheres with the isotropic radiator at the center. If the antenna is designed to concentrate a majority of its energy in one or more directions, it is said to be directional. Since the directional antenna radiates the same total power as it would if it were an isotropic radiator, gain exists in the direction(s) of power concentration. That gain is measured in decibels relative to an isotropic radiator (dBi).
7. What is the power dynamic range of an ideal 12-bit analog-to-digital converter (ADC)?
c) 72.25 dB
An ideal 12-bit ADC can assume 212 (4,096) unique voltage levels. Since power is proportional to the square of the voltage, the maximum power sample value is 40962 (16,777,216) times the minimum power sample value. Therefore the dynamic range is 10*log (16,777,216) = 72.25 dB.
A rule of thumb is 6 dB per bit.
8. An ideal 10 dB attenuator is added in front of a load that has a 2.00:1 VSWR. What is the resulting VSWR of the load + attenuator?
a) 1.07:1
VSWR is related to return loss (RL) according to VSWR = [10^(RL/20) + 1] / [10^(RL/20) - 1]. It follows that increasing the return loss will result in a lower VSWR. The RL of a 2.00:1 VSWR is 9.542 dB. Add the 10 dB attenuator for a total RL of 2*10 dB + 9.542 dB = 29.542 dB. Convert back to VSWR using the given formula for a value of 1.07:1.
Why add twice the attenuator value to the return loss? Return loss is the total decrease in signal strength in passing through the attenuator and being reflected back through the attenuator. Hence, the signal is decreased by twice the attenuator value.
9. What is the thermal noise power in a 1 MHz bandwidth when the system temperature is 15 °C (assume gain and noise figure are 0 dB)?
a) -114.0 dBm (in a 1 MHz BW)
Thermal noise power density is governed by the equation 10*log (k*T*B*1000) dBm, where k is the Boltzmann constant. T is the temperature in degrees Kelvin, and B is the bandwidth in Hertz. Multiplication by 1000 is to convert watts to milliwatts. A rule of thumb for temperatures near 15 °C is to begin with a thermal noise density of -174 dBm/Hz, and scale accordingly (add 10 dB per decade of increased bandwidth).
10. Two equal amplitude tones have a power of +10 dBm, and generate a pair of equal amplitude 3rd-order intermodulation products at -20 dBm. What is the 2-tone, 3rd-order intercept point (IP3) of the system?
b) +25 dBm
2-tone, 3rd-order intermod products increase 3 dB in power for every 1 dB increase in tones that produce them. That means the intermods increase in power at a rate of 2 dB per 1 dB relative to the tone power. The 2-tone, 3rd-order intercept point is defined as the theoretical point where the two original tones and the two 3-rd-order products would have equal power (not possible in real systems due to saturation limits).
If the two original tones have a power of +10 dBm and the 3rd-order products have a power of -20 dBm, then the intercept point will be at +10 dBm + [(+10) - (-20)]/2 dB = +10 dBm + 15 dB = +25 dBm.

awp666

1.b
2.b
3.Not sure
4.b
5.b
6.c
7.??
8.a
9.a
10.??

awp666

60分,还要努力啊~~~~[em10]

checkz

定向耦合器是一种四端口网络，
定向耦合器是无源和可逆网络。理论上，定向耦合器是无耗电路，而且其各个端口均应是匹配的。图１（ｂ）定义了定向耦合器各端口的属性。当信号从端口１输入时，大部分信号从端口２直通输出，其中一小部分信号从端口３耦合出来，端口４通常接一个匹配负载。如果要将定向耦合器反过来使用，则端口１和２，端口３和４的属性要互换定义。
定向耦合器可以由同轴、波导、微带和带状线电路构成。通常，定向耦合器用于信号取样以进行测量和监测，信号分配及合成；此外，作为网络分析仪，天线分析仪和通过式（ＴＨＲＵＬＩＮＥ?）功率计等测试仪器的核心部件，定向耦合器所起的作用是正向和反射信号的取样。定向耦合器的方向性是一项至关重要的指标，尤其是作为信号合成和反射测量应用时。

２． 各项指标的定义
如图１（ｂ）所示，在理想情况下，当信号功率从端口１输入时，输出功率只应出现在端口２和端口３，而端口４是完全隔离的，没有功率输出。但是在实际情况下，总有一些功率会泄漏到端口４。设端口１的输入功率为Ｐ１，端口２，３和４的输出功率分别为Ｐ２，Ｐ３和Ｐ４，则定向耦合器的特性可以由耦合度，插入损耗，隔离度和方向性等四项指标来表征，单位均为ｄＢ。
请注意在以下的描述中，所有的指标均表示为正数，而在实际应用中，则是用负数来进行各种计算的。
耦合度：
耦合度表示从端口１输入的功率和被耦合到端口３部分的比值，表示为：耦合度（Ｃ）＝１０×ｌｏｇ（Ｐ１?Ｐ３）
插入损耗：
插入损耗表示从端口１到端口２的能量损耗，表示为：插入损耗（ＩＬ）＝１０×ｌｏｇ（Ｐ１?Ｐ２）
请注意端口１的输入功率有一部分功率是被耦合到端口３的，所以应导入一个“耦合损耗”的概念，下面是各种耦合度下的耦合损耗值：
耦合度 耦合损耗
６ ｄＢ １．２００ ｄＢ
１０ ｄＢ ０．４６０ ｄＢ
１５ ｄＢ ０．１４０ ｄＢ
２０ ｄＢ ０．０４０ ｄＢ
３０ ｄＢ ０．００４ ｄＢ
通常所说的从端口１到端口２的插入损耗是传输损耗和耦合损耗之和。在定向耦合器的产品说明中通常会对此加以特别说明。
当定向耦合器用于测试和测量时，选取的耦合度比较小，如２０ｄＢ或３０ｄＢ甚至更小；而作为功率合成系统或者信号分配系统应用时，则会采用比较大的耦合度，如３ｄＢ，５ｄＢ和７ｄＢ等。
隔离度：
前面提到，在理想的定向耦合器中，端口４是没有功率输出的，而实际上总会有一些功率从这个端口泄漏出来，这就是隔离度的指标，表示为：隔离度（Ｉ）＝ １０×ｌｏｇ（Ｐ１?Ｐ４）
方向性：
端口３的输出功率和端口４输出功率之间的比值定义为方向性，表示为：方向性（Ｄ）＝１０×ｌｏｇ（Ｐ３?Ｐ４）
需要特别说明的是耦合度，隔离度和方向性之间的关系为：
隔离度（Ｉ）＝耦合度（Ｃ）＋方向性（Ｄ）
耦合度是一项设计指标，是根据使用要求而选定的，通常为６，１０，２０和３０ｄＢ，这样隔离度指标也随之而变化；而方向性则是一个常数。
在大部分定向耦合器的指标中，通常只标出方向性指标，隔离度指标可以根据耦合度计算出来。如：
耦合度（Ｃ）＝３０ｄＢ，
方向性（Ｄ）＝２５ｄＢ，
则隔离度（Ｉ）＝３０＋２５＝５５（ｄＢ）

<P align=right><FONT color=red>+5 RD币</FONT></P>

<P align=right><FONT color=red>+5 RD币</FONT></P>

感谢checkz的考题，连同前几贴一同加分。期待后续系列啊！！！
[em08][br]<p align=right><font color=red>+5 RD币</font></p>

checkz

来一个smith chart的知识 每次面试总被问到呵呵
【文件名】:0669@52RD_smithchart.rar
【格　式】:rar
【大　小】:508K
【简　介】:
【目　录】:
[br]<p align=right><font color=red>+1 RD币</font></p>[br]<p align=right><font color=red>+1 RD币</font></p>

checkz

多谢lz鼓励，我会继续努力的，

lanneret

<P>一半懂一半不懂！看来还要好好进修啊！！！！！</P>[em01]

lanneret

<P>继续啊，期待中！长进不少！！！！！这些题目哪里来的。</P>

awp666

<P>期待楼主的新考题啊～～～</P>[em01]

checkz

1. What is a primary advantage to using 90 degree (quadrature) hybrid couplers in amplifier designs?
a) Wider bandwidth possible
b) Low noise figure
c) Input/output impedance not dependent on devices as long as device impedances are equal
2. Why is there a frequency term in the equation for free-space path loss?
a) There is no frequency term
b) Atmospheric absorption
c) Antenna geometry requires it
3. If an amplifier has a noise temperature of 60K, what is its noise figure for an ambient temperature of 290K?
a) 8.0 dB
b) 80 dB
c) 0.82 dB
4. What is a primary advantage of offset-quadrature-phase-shift-keying (OQPSK) over standard QPSK?
a) Greater data rates possible
b) Greater spectral efficiency
c) More constant envelope power
5. A mixer has the following input frequencies:  RF = 800 MHz, LO = 870 MHz. The desired output frequency is 70 MHz. What is the image frequency?
a) 940 MHz
b) 1670 MHz
c) 140 MHz
6. What is the spurious-free dynamic range of a system with IP3 = +30 dBm and a minimum discernible signal (MDS) level of -90 dBm?
a) 80 dB
b) 120 dB
c) 60 dB
7. A spectrum analyzer displays a component at 10 MHz @ 0 dBm, 30 MHz @ -10 dBm, 50 MHz @ -14 dBm, 70 MHz @ -17 dBm, and all of the other odd harmonics until they disappear into the noise. What was the most likely input signal that caused the spectrum?
a) A 10 MHz square wave (0 Vdc)
b) A 10 MHz triangle wave (0 Vdc)
c) A 10 MHz cosine wave (0 Vdc)
8. On which side of a rectangular waveguide is an E-bend made?
a) The long dimension
b) The short dimension
c) The inside
9. During a network analyzer calibration, why are both a short circuit and an open circuit used?
a) They average to 50 ohms in an RF system
b) To determine the characteristic impedance of the measurement system
c) Both are easy to produce to at high accuracy
10. What is the first harmonic of 1 GHz?
a) 1 GHz
b) 2 GHz
c) 10 GHz


大家继续努力哟....